Proof about the subgroup of injective functions being the subgroup of group of even permutations I have the following problem:
Let $H$ be the subgroup of $S_n$, where $S_n$ denotes the group of injective functions mapping a set to itself, and the order of $H$ is odd. I need to prove that $H$ is a subgroup of $A_n$ where $A_n$ is an alternating group of degree n.
My idea: If the order of $H$ is odd, then it means that the least common multiple of length of the disjoint cycles that make up $H$ is odd, which implies that there are only cycles of odd length. I think all cycles can be written as a multiple of 2-cycles, thus, I can conclude that $H$ is a subgroup of the group of even permutations?
Am I right? Any help would be appreciated!
Thanks in advance!
 A: In $S_n$, every permutation is either even or odd, right? And the order (the size) of $A_n$ is exactly half the size of $S_n$. You should already know why this is true.
Now look at even and odd permutations inside $H$, that is, consider $H\cap A_n$ which is a subgroup of $H$. You want to prove that it is in fact all of $H$. What if it's not? Then $H$ has both even and odd permutations. By analogy with all of $S_n$, can you try to prove that in that case inside $H$ the number of even and odd permutations is the same? If you prove that, do you see then how this establishes the contradiction you need?
A: Since $A_n$ is a normal subgroup of $S_n$, we have that $HA_n$ is a subgroup of $S_n$; in case $H\not\subseteq A_n$, this subgroup is the whole of $S_n$, because $A_n$ is maximal, and so
$$
S_n/A_n=(HA_n)/A_n\cong H/(H\cap A_n)
$$
by the homomorphisms theorems.
But $|S_n/A_n|=2$, contradiction.

You're confusing things: a permutation can be written as a product of disjoint cycles, but you can't speak about “the disjoint cycles that make up $H$”.
A: Have you seen in your course (or maybe an earlier one) that if $AB$ are subgroups of a finite group $G$, then the set $AB$ (which need not be a subgroup in general) has cardinality $\frac{|A||B|}{|A \cap B|}$. If you already know that $[S_{n}:A_{n}] = 2$, you can then check that when $H$ is a subgroup of $S_{n}$ of odd order, the set $HA_{n}$ has cardinality $[H :H \cap A_{n}]|A_{n}|$, which is an odd integer multiple of $|A_{n}|$. Since $|S_{n}| = 2|A_{n}|$, we can only have $|HA_{n}| = |A_{n}|,$ so that $H \subseteq A_{n}$.
Alternatively, you can proceed along the lines you were thinking: any $k$-cycle in $S_{n}$ may be written as a product of $k-1$ $2$-cycles. Hence when $k$ is odd, any $k$-cycle is a even permutation, and lies in $A_{n}$. If $H$ has odd order, as you came close to saying, it is the case that every element $h \in H$ can be written as  a product of disjoint cycles, all of odd length, and every such $h$ lies in $A_{n}$.
